Barying Pascal

by v_Enhance, Nov 4, 2012, 6:43 PM

Let $A=(1,0,0)$, $C=(0,1,0)$ and $E=(0,0,1)$. Set
\begin{align*}
	D &= (1 : y_2 : z_2) \\
	F &= (x_3 : 1 : z_3) \\
	B &= (x_1 : y_1 : 1)
\end{align*}

Because the points $D$, $F$, $B$ lie on the $(ACE) : a^2yz + c^2zx + e^2xy = 0$, we obtain \[  \det \left( \begin{array}{ccc} y_2z_2 & z_2 & y_2 \\ z_3 & z_3x_3 & x_3 \\ y_1 & x_1 & x_1y_1 \end{array} \right) = 0. \]
since the columns are linearly dependent.


Now $BC \cap EF = ( y_1 : y_1y_2 : y_2)$, et cetera. Then, compute \[ \det \left( \begin{array}{ccc} x_3x_1 & x_1 & x_3 \\ y_1 & y_1y_2 & y_2 \\ z_3 & z_2  & z_2z_3 \end{array} \right) = \det \left( \begin{array}{ccc} y_2z_2 & z_2 & y_2 \\ z_3 & z_3x_3 & x_3 \\ y_1 & x_1 & x_1y_1 \end{array} \right) = 0. \]
So $BC \cap EF$, $CD \cap FA$, and $DA \cap AB$ are collinear.
This post has been edited 2 times. Last edited by v_Enhance, Mar 22, 2015, 4:49 PM
Reason: Unbreak LaTeX

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2 Comments

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not bad, .

by ksun48, Nov 4, 2012, 9:43 PM

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wait pretty much all projective theorems are baryable since bary $\in$ projective coordinates.

by thkim1011, Nov 15, 2014, 6:39 PM

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